circuit analysisresistors
Resistors R2,R3,R4 are connected in series. I'm not sure how to find the equivalent resistance of between terminals A and B, because of the short circuit. However, I suspect the answer is R1+R4 because of the short circuit. I am confused about this, can someone help me? Thanks!
EDIT: Thanks for pointing that out. My drawing is indeed incorrect. Lets say the resistor R4 which is connected in series to R2, R3 will now be called R5
There are two steps to finding the Thevenin equivalent circuit: finding the Thevenin voltage and the Thevenin resistance.
Thevenin voltage is the voltage across the two points you interested in (Vin). In this case it is easy to calculate as there is no current flowing in the 43 and 60 \$\Omega\$ resistors thus no voltage drop. Thus the voltage at Vin is the same as the voltage form the source, 72 V.
Thevenin resistance is calculated by 'turning off' all independent current and independent voltage sources and calculating the resistance between the two points. Turning off a voltage source sets the voltage across it to 0, which results in a short (0 \$\Omega\$) in parallel with the 275 \$\Omega\$ resistor. Any resistor combined in parallel with a short results in a short, leaving you with the 43 and 60 \$\Omega\$ resistors now in series, giving a Thevenin resistance of 103 ohms.
Putting the two together gives you a voltage source of 72 V in series with a 103 \$\Omega\$ resistor for you Thevenin equivalent circuit.
That circuit can be simplified only to a point by using only series/parallel transformations. By using those you end up with this circuit:
where \$R12 = R1 + R2 \$, \$R34 = R3 + R4\$ and \$Rs = R8 + R9 + R10\$.
As you can see no further reduction can be done with series or parallel transformations.
At this point you have two possibilities:
Apply a voltage source to the terminals, calculate (using mesh analysis, for example) the current drawn by the circuit and calculate the equivalent resistance as V/I.
Simplify the circuit further applying the so called star/delta transformations.
With the second approach you need in particular these formulas to transform, for example, the delta connection of R5, R6 and R7 into a star connection. Once you do this, the circuit can be further simplified using usual series/parallel transformations.
Best Answer
If your drawing is correct, then the resistance between A & B is simply R1 + R4. R2, R32 and R4 are not relevant since they are shorted out. Note, however, that there are 2 resistors with the same label, R4.